question_answer

Consider a system of two particles having masses \[{{m}_{1}}\] and \[{{m}_{2}}\]. If the particle of mass \[{{m}_{1}}\]is pushed towards the mass centre of particles through a distance d, by what distance would the particle of mass \[{{m}_{2}}\]move so as to keep the mass centre of particles at the original position?

A) \[\frac{{{m}_{1}}}{{{m}_{1}}+{{m}_{2}}}d\]                         

B) \[\frac{{{m}_{1}}}{{{m}_{2}}}d\]

C) d                                            

D) \[\frac{{{m}_{2}}}{{{m}_{1}}}d\]

Correct Answer: B

Solution :

The system of two given particles of masses \[{{m}_{1}}\] and \[{{m}_{2}}\] are shown in figure. Initially the centre of mass\[{{r}_{GM}}=\frac{{{m}_{1}}{{r}_{1}}+{{m}_{2}}{{r}_{2}}}{{{m}_{1}}+{{m}_{2}}}\].(i) When mass \[{{m}_{1}}\]moves towards centre of mass by a distance d, then let mass \[{{m}_{2}}\] moves a distance d' away from CM to keep the CM in its initial position. So\[{{r}_{CM}}=\frac{{{m}_{1}}({{r}_{1}}-d)+{{m}_{2}}({{r}_{2}}+d')}{{{m}_{1}}+{{m}_{2}}}\]                       ?(ii) Equating Eqs. (i) and (ii), we get \[\frac{{{m}_{1}}{{r}_{1}}+{{m}_{2}}{{r}_{2}}}{{{m}_{1}}+{{m}_{2}}}=\frac{{{m}_{1}}({{r}_{1}}-d)+{{m}_{2}}({{r}_{2}}+d')}{{{m}_{1}}+{{m}_{2}}}\] \[\Rightarrow \]\[-{{m}_{1}}d+{{m}_{2}}d'=0\Rightarrow d'=\frac{{{m}_{1}}}{{{m}_{2}}}d\]